80. Find both the maximum value and the minimum value of 3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3].

We have, f (x) = 3x4- 8x3 + 12x2- 48x + 25, x ∈ [0, 3].f (x) = 12x3- 24x2 + 24x - 48.At f (x) = 0.12x3- 24x2 + 24x - 48 = 0.x3- 2x2 + 2x - 4 = 0x2 (x - 2) + 2 (x - 2) = 0 (x - 2) + (x2 + 2) = 0x = 2 ∈ [0, 3] or x = ±√-2 which is not possible as∴f (x) = 3 (2)4- 8 (2)3 + 12 (2)2- 4 (2) + 25.=48 - 64 + 48 - 96 + 25.= -39.f (0) =3 (0)4- 8 (0)3 + 12 (0)2- 48 (0) + 25.= 25.f (3) = 3 (3)4- 8 (3)3 + 12 (3)2- 48 (3) + 25.= 243 - 216 + 108 - 144 + 25= 16.Maximum value of f (x) = 25 at x = 0.and minimum value of f (x) = -39 at x = 2.

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Ncert Solutions Maths class 12th Maths Class 12th Application of Derivatives

80. Find both the maximum value and the minimum value of 3x⁴ – 8x³ + 12x² – 48x + 25 on the interval [0, 3].

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Vishal Baghel | Contributor-Level 10

4 months ago

We have, f (x) = 3x⁴- 8x³ + 12x²- 48x + 25, x ∈ [0, 3].

f (x) = 12x³- 24x² + 24x - 48.

At f (x) = 0.

12x³- 24x² + 24x - 48 = 0.

x³- 2x² + 2x - 4 = 0

x² (x - 2) + 2 (x - 2) = 0

(x - 2) + (x² + 2) = 0

x = 2 ∈ [0, 3] or x = $\pm\sqrt-2$ which is not possible as

∴f (x) = 3 (2)⁴- 8 (2)³ + 12 (2)²- 4 (2) + 25.

=48 - 64 + 48 - 96 + 25.

= -39.

f (0) =3 (0)⁴- 8 (0)³ + 12 (0)²- 48 (0) + 25.

= 25.

f (3) = 3 (3)⁴- 8 (3)³ + 12 (3)²- 48 (3) + 25.

= 243 - 216 + 108 - 144 + 25

= 16.

Maximum value of f (x) = 25 at x = 0.

and minimum value of f (x) = -39 at x = 2.

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